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3 years ago

What is the differences between 2^8 and 8^2?

Mathematics

2 answers:

Fynjy0 [20]3 years ago

7 0

Answer:

The answer is 192

Step-by-step explanation:

2^8 = 256

8^2 = 64

256 - 64 = 192

Remember:

2^8 is really 2*2*2*2*2*2*2*2

Remember:

8^2 is really 8*8

To find the <u>difference,</u> you must subtract the smaller product from the larger product.

dsp733 years ago

4 0

Answer:

192

Step-by-step explanation:

Method 1:

2^8 - 8^2 = 2^8 - 2^6 = 2^6(2^2 - 1) = 64(3) = 192

Method 2:

2^8 - 8^2 = 256 - 64 = 192

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Write each equation in standard form. y=-2/3x+3

GREYUIT [131]

Answer:

nvm i forgot

Step-by-step explanation:

8 0

3 years ago

Please help me. I suck at math

serg [7]

Answer:

Answer is ...Obtuse

Step-by-step explanation:

$\frac{m$

$\frac{5x + 4}{2} = \frac{3}{2} x + 21$

$\frac{5x + 4}{2} = \frac{3x + 42}{2}$

$5x + 4 = 3x + 42$

$5x - 3x = 42 - 4$

$2x = 38$

$x = 19$

$substituting \: x = 19 \: \:in \: \: \: \: \: \: \: \\ m$

We get,

m<ABC =

$(5 \times 19) + 4 = 99$

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4 0

3 years ago

How do you find the derivative of 1/logx?

valentina_108 [34]

The derivative of 1/logx is With the chain rule.

1log(x)=log(x)−1 is ,= -1xlog(x)2 .

The by-product of logₐ x (log x with base a) is 1/(x ln a). Here, the thrilling issue is that we have "ln" withinside the by-product of "log x". Note that "ln" is referred to as the logarithm (or) it's miles a logarithm with base "e".

The by-product of 1/log x is -1/x(log x)^2. Note that 1/logx is the reciprocal of log.

$With the chain rule. \\ 1log(x)=log(x)−1 \\ So you follow the chain rule to the electricity after which to the log: \\ ddxlog(x)−1=−1⋅log(x) \\ \\−2⋅ddxlog(x) \\ =−1log(x)21x \\ =−1xlog(x)2 .$